/sci/ - Science & Math

[Anonymous No.2173333]

Feel free to publish this.

[Anonymous No.2173337]

middle term is not a number in the decimal number system

[Anonymous No.2173338]

Your first two terms are swapped.

0.9...99 > 0.9...91

[Anonymous No.2173343]

you cannot have 0.9 recurring followed by a 1, otherwise that's not 9 recurring is it, you do not understand the concepts involved, you retard.

[Anonymous No.2173341]

0.9...1
isnt even a number

[Anonymous No.2173357]

In every decimal number, you can replace a digit with an alternate value and produce a different number.

[Anonymous No.2173367]

Guy on the pic's wrong. The second term does not exist. There can't be a 1 after the last digit as there is no last digit in a recurring decimal. And since the difference between 0.99 recurring and 1.0 recurring converges to 0 there is theoretically no difference. 2/3 = 0.66 recurring, 1/3 = 0.33 recurring. 0.33 recurring + 0.66 recurring = 0.99 recurring, 2/3 + 1/3 = 1 => 0.99 recurring = 1.

PUBLISH THAT!

[Anonymous No.2173375]

>>2173357

so long as that digit resides at a finite location.

.9...1 is saying put a 1 at the infinite digit, which isnt possible.

You could perhaps do a limit, where you see what happens as you placed the nth digit as n goes to infinity, in which case, .9...1 = .9... as n goes to infinity.

[Anonymous No.2173377]

>Divde 1 by 998999
>Receive Fibonacci numbers
>???
>http://www.youtube.com/watch?v=wCNICT2en_U&feature=related

[Anonymous No.2173380]

>>2173357
true, but every individual digit is at a FINITE number of decimal places from the decimal point

so replacing would produce

0.999999999999999999919....

or 0.999999999999999999999999999919... etc

[dolan !Uq56cE.45s No.2173393]

>>2173377

[Anonymous No.2173407]

>>2173377
>>2173393
I think you get the same thing by dividing by 89 or 9899, just, the numbers are superimposed.

[Anonymous No.2173432]

>>2173377

[Anonymous No.2173435]

>>2173407
omfg, now I can have all of the 4 digit, 5 digit etc. fibonacci numbers!
1/99989999 gives it me as 1.00010002... which will allow for the 4 digit ones.
1/9999899999 gives me the 5 digit ones.

WOAAAAH-OH-OH-AHHHHH

[Anonymous No.2173453]

>>2173407
so this number basically works because we use the base 10 system, so for n-digit numbers we use

1/(10^2n - 10^n -10^0)

does that mean that if we use base b, the expression becomes:
1/(b^2n - b^n - b^0)

?

[Anonymous No.2173464]

>>2173453
oh snap oh snap

if we introduce another term, i.e.

1/(10^3n - 10^2n - 10^n - 10)

we get numbers that are like fibonacci numbers, but are the sum of the previous THREE numbers instead of the previous two!

[Anonymous No.2173468]

Oh nO OP discovered decimall hyperreals

[Anonymous No.2173471]

Op trolled 9 people so far

1 point for each op

9/10

[Anonymous No.2173486]

<span class="math">\displaystyle 0.999 \dots= \lim_{n\to \infty} 0.\underbrace{99 \dots 9\,}_{n} = \lim_{n\to \infty} \sum_{k=1}^n \frac{9}{10^k} = \lim_{n\to \infty} (1 - \frac{1}{10^n}) = 1 - \lim_{n\to \infty} \frac{1}{10^n} = 1[/spoiler]

Habeeb it. If you do not understand this notation then you don't even have a saying on the matter at hand.

[Anonymous No.2173497]

>>2173486
>mfw I understood everything

[Anonymous No.2173495]

>>2173464
my bad, that should be

1/(10^3n - 10^2n - 10^n - 10^0)

[Anonymous No.2173506]

>>2173468
don't even bother, no one in this thread even knows what hyper-reals are. which is kind of sad, because the hyper-reals can singlehandedly stop all the troll math on sci.

[Anonymous No.2173518]

working with hyperreals
0.9... = 1 - dx < 1
so
0.999... < 1

but st(0.999...) = st(1 - dx) = 1
so in the reals 0.999... = 1

[Anonymous No.2173520]

>>2173506

[Anonymous No.2173528]

>>2173506
i understand, it's just non standard analysis is unnecessary faggotry, albeit logically sound unnecessary faggotry.

standard analysis, if you will, is good enough. shame so few understand it

[Anonymous No.2173537]

>>2173528

<span class="math">/displaystyle \begin{align}
0{,}999\dots &= 9 \cdot 10^{-1} + 9 \cdot 10^{-2} + 9 \cdot 10^{-3} + \dots \\
&= 9 \cdot (10^{-1} + 10^{-2} + 10^{-3} + \dots) \\
&= 9 \cdot \sum_{n=1}^{\infty} \frac{1}{10^n} \\
&= 9 \cdot \frac{\frac{1}{10}}{1-\frac{1}{10}} = 9 \cdot\frac{1}{9} = 1.
\end{align}[/spoiler]

I trust the proof of a geometric series sum is not necessary here.

[Anonymous No.2173547]

>>2173506
>the hyper-reals can singlehandedly stop all the troll math on sci.

I doubt that.

[Anonymous No.2173557]

>>2173528

How about:

<span class="math">/displaystyle 0.999\dots = 9 \cdot 10^{-1} + 9 \cdot 10^{-2} + 9 \cdot 10^{-3} + \dots[/spoiler]
<span class="math">\displaystyle = 9 \cdot (10^{-1} + 10^{-2} + 10^{-3} + \dots)[/spoiler]
<span class="math">\displaystyle = 9 \cdot \sum_{n=1}^{\infty} \frac{1}{10^n}[/spoiler]
<span class="math">\displaystyle = 9 \cdot \frac{\frac{1}{10}}{1-\frac{1}{10}} = 9 \cdot\frac{1}{9} = 1.[/spoiler]

I trust that the proof of the geometric series sum is not in order.

[Anonymous No.2173563]

>>2173547
all the troll math (/0, 1 = 2 (by using /0), infinity related, 0.999... = 1) can be easily untroled with hyperreals

[Anonymous No.2173564]

>>2173528

Standard it is then

<span class="math">\displaystyle 0.999\dots = 9 \cdot 10^{-1} + 9 \cdot 10^{-2} + 9 \cdot 10^{-3} + \dots[/spoiler]
<span class="math">\displaystyle = 9 \cdot (10^{-1} + 10^{-2} + 10^{-3} + \dots)[/spoiler]
<span class="math">\displaystyle = 9 \cdot \sum_{n=1}^{\infty} \frac{1}{10^n}[/spoiler]

<span class="math">\displaystyle = 9 \cdot \frac{\frac{1}{10}}{1-\frac{1}{10}} = 9 \cdot\frac{1}{9} = 1.[/spoiler]

I trust that the proof of the geometric series sum is not in order.

[Anonymous No.2173562]

>>2173557
how does i put pictures in my text?

[Anonymous No.2173595]

>>2173528
IIRC The real numbers with the usual sum, product and inequalities are a elementary submodel of the hyperreal numbers, as a consequence you wont get new theorems from the real numbers by working with hyperreals.

Also for the construction of the hyperreals you need an ultrafilter on the natural numbers. Trivial ultrafilters will give trivial sets (something isomorphic to the real numbers), so you need a free ultrafilter. You need axiom of choice for that.
Also stuff like .99...1 doesn't make sense in the hyperreals.

[Anonymous No.2173632]

>>2173595
let x,y be infintismals
0.999...1 = (1-x)+y

[Anonymous No.2173631]

>>2173595

0.99...1 doesn't make sense anywhere, it's not legit notation.

What they are going for should always be specified

<span class="math">\displaystyle 0.999\underbrace{\dots}_{n}1 , n ≠ \infty[/spoiler]

[Anonymous No.2173656]

>>2173595

0.99...1 doesn't make sense anywhere; it's not legit notation.

What most people are going for with this should always be specified with

<span class="math">\displaystyle 0.99\underbrace{\dots}_{n}1[/spoiler]
<span class="math">\displaystyle n ≠ \pm\infty[/spoiler]

[Anonymous No.2173677]

>>2173656
What a terrible definition. Anybody would expect that .9..1 is unique.

>>2173632
If you're with w+1 sequences then it does not make sense. Hyperreals are constructed by a quotient with w-sequences of real numbers.

[Anonymous No.2173680]

>>2173677
*If you're working with w+1 sequences then it does make sense.

[Anonymous No.2173698]

fucking hell. DON"T RESPOND to this shit. If you do, at least use sage.

[Anonymous No.2173794]

>>2173656
>>2173564
what type of black magic are you performing to write like that?